Question: Consider the parametric curve: $\begin{aligned} x&=5t^2 \\\\ y&=2e^t \end{aligned}$ Which integral gives the arc length of the curve over the interval from $t=-1$ to $t=3$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_{-1}^3 \sqrt{100t^2+4e^{2t}}\,dt$ (Choice B) B $\int_{-1}^3 \sqrt{5t^2+2e^{t}}\,dt$ (Choice C) C $\int_{-1}^3 \sqrt{10t+2e^{t}}\,dt$ (Choice D) D $\int_{-1}^3 \sqrt{25t^4+4e^{2t}}\,dt$
Explanation: This is the formula for the arc length of a parametric curve over the interval $[a, b]$ : $\int_a^b\sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\, dt$ [Where does this formula come from?] Let's find $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{d}{dt}\left[5t^2\right] \\\\ &=10t \\\\\\ \dfrac{dy}{dt}&=\dfrac{d}{dt}\left[2e^t\right] \\\\ &=2e^t \end{aligned}$ Now we can find the expression for the arc length: $\begin{aligned} &\phantom{=}\int_{-1}^3 \sqrt{\left(\dfrac{dx}{dt}\right)^2+\left(\dfrac{dy}{dt}\right)^2}\,dt \\\\ &=\int_{-1}^3 \sqrt{(10t)^2+(2e^t)^2}\,dt \\\\ &=\int_{-1}^3 \sqrt{100t^2+4e^{2t}}\,dt \end{aligned}$ In conclusion, this integral gives the arc length of the curve over the interval from $t=-1$ to $t=3$ : $\int_{-1}^3 \sqrt{100t^2+4e^{2t}}\,dt$